Quadratic equations question

Question

let $P(x)$ and $Q(x)$ be 2 quadratic eqs. with one non-rational root common and integral coefficients. prove $P(x) = r.Q(x)$, for some rational no. $r$

TRIED ANSWER:

Let $P(x)= ax^2 + bx + c$

$\Rightarrow$ $x = \frac{-b \pm \surd b^2-4ac}{2a}$

Let $Q(x)= mx^2 + nx + o$

$\Rightarrow$ $x' = \frac{-n \pm \surd n^2-4mo}{2m}$

If one root of $P(x)$ = one root of $Q(x)$

then assume $\Rightarrow$ $\frac{-b + \surd b^2-4ac}{2a}$ = $\frac{-n + \surd n^2-4mo}{2m}$

As they both are irrational, comparing the two we get the irrational part equal.

Hence both the roots are same for the two equations

OR

They are the same equations with different leading coefficients.

THUS

$P(x)=k*Q(x)$

for some constant k.

Answer 1

The statement is false. For example, $P(x)=(x-\sqrt 2) = x^2 - 2\sqrt 2 x + 2$ and $Q(x) = x^2-2$ have one non-rational root in common. However, there is no rational number $r$ such that $P(x) = r\cdot Q(x)$.

Answer 2

Let $P(x)=ax^2+bx+c$ and $Q(x) = dx^2+ex+f$. Let the roots of $P(x)$ be $\alpha,\beta$ and of $Q(x) $ be $\alpha,\gamma$. $\alpha$ is the common irrational root.
$\alpha+\beta=\frac{-b}a$. Since $-\frac ba$ is irrational, $\alpha+\beta$ must also be irrational. $\implies\beta=-\alpha$. $\implies -\frac ba=0 \implies b = 0$
Similarly, $\gamma = -\alpha$ and $e=0$.

$\alpha\cdot\beta=\frac ca \implies-\alpha^2=\frac ca = \frac fd$.

$P(x)=ax^2+c$ and $Q(x)=dx^2+f$.
$P(x)=ax^2-a\cdot\alpha^2$ and $Q(x)=dx^2-d\alpha^2$.
$P(x)=ax^2-a\cdot\alpha^2=\frac ad(dx^2-d\cdot\alpha^2)=\frac adQ(x)=r\cdot Q(x)$.